Let x = 9 be a directrix of an ellipse E , whose centre is at the...

Given the directrix of the ellipse is $x = \dfrac{a}{e} = 9$ and eccentricity $e = \dfrac{1}{3}$.

$\dfrac{a}{1/3} = 9 \Rightarrow a = 3$

The value of $b^2$ is given by $b^2 = a^2(1 - e^2) = 9\left(1 - \dfrac{1}{9}\right) = 8$.

The equation of the ellipse is $\dfrac{x^2}{9} + \dfrac{y^2}{8} = 1$.

The focus $P(\alpha, 0)$ for $\alpha > 0$ is at $(ae, 0) = \left(3 \times \dfrac{1}{3}, 0\right) = (1, 0)$.

Let the midpoint of the chord $AB$ be $(h, k)$. The equation of the chord in terms of its midpoint is given by $T = S_1$ :

$\dfrac{hx}{9} + \dfrac{ky}{8} = \dfrac{h^2}{9} + \dfrac{k^2}{8}$

Since the chord passes through the focus $P(1, 0)$, substituting $x = 1$ and $y = 0$ gives:

$\dfrac{h}{9} = \dfrac{h^2}{9} + \dfrac{k^2}{8}$

$\dfrac{k^2}{8} = \dfrac{h}{9} - \dfrac{h^2}{9}$

$9k^2 = 8h(1 - h)$

Replacing $(h, k)$ with $(x, y)$, the locus of the midpoint is:

$9y^2 = 8x(1 - x)$

Answer: $9y^2 = 8x(1-x)$